Question

(i) If the vertices of a triangle are (1, −3), (4, p) and (−9, 7) and its area is 15 sq. units, find the value(s) of p.

(ii) Find the value of k so that the area of triangle ABC with A(k + 1, 1), B(4, –3) and C(7, –k) is 6 square units.

Answer 1

(i) Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC.

Here, x₁ = 1, y₁ = −3; x₂ = 4, y₂ = p and x₃ = −9, y₃ = 7

ar(∆ABC) = 15 square units

$$\begin{gathered} \Rightarrow \frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|=15 \\ \Rightarrow \frac{1}{2}\left|1\left(p-7\right)+4\left[7-\left(-3\right)\right]+\left(-9\right)\left(-3-p\right)\right|=15 \\ \Rightarrow \frac{1}{2}\left|p-7+40+27+9p\right|=15 \\ \Rightarrow \left|10p+60\right|=30 \end{gathered}$$

$\Rightarrow 10p+60=30$ or $10p+60=-30$

$\Rightarrow 10p=-30$ or $10p=-90$

$\Rightarrow p=-3$ or $p=-9$

Hence, the value of p is −3 or −9.

(ii) Area of the triangle formed by the vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)\mathrm{is}$ $\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$.

Now, the given vertices are A(k + 1, 1), B(4, –3) and C(7, –k)
and the given area is 6 square units.

Therefore,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|\left(k+1\right)\left(-3-\left(-k\right)\right)+\left(4\right)\left(-k-1\right)+\left(7\right)\left(1-\left(-3\right)\right)\right| \\ \Rightarrow 6=\frac{1}{2}\left|\left(k+1\right)\left(-3+k\right)+\left(4\right)\left(-k-1\right)+\left(7\right)\left(1+3\right)\right| \\ \Rightarrow 6=\frac{1}{2}\left|-3k+k^2-3+k-4k-4+\left(7\right)\left(4\right)\right| \\ \Rightarrow 6=\frac{1}{2}\left|-3k+k^2-3+k-4k-4+28\right| \\ \Rightarrow 12=\left|k^2-6k+21\right| \\ \Rightarrow k^2-6k+21=12 \\ \Rightarrow k^2-6k+21-12=0 \\ \Rightarrow k^2-6k+9=0 \\ \Rightarrow {\left(k-3\right)}^2=0 \\ \Rightarrow k-3=0 \\ \Rightarrow k=3 \end{gathered}$$

Hence, ​the value of k is 3.